Linked Questions
11 questions linked to/from Is the Momentum Operator a Postulate?
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Why is quantum mechanical momentum the derivative of the wave function with respect to the position? [duplicate]
In classical mechanics the momentum is defined as mass times the time-derivative of position.
In quantum mechanics, however, the time-derivative of the wave function is the hamiltonian, while the ...
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Common observables and associated operators: operator momentum [duplicate]
Starting from my previous question Commutators in quantum mechanics and considering that the commutator
$$\left[i\hbar\frac{\partial}{\partial x},x\right]=i\hbar, \tag{1}$$
the associated linear ...
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Does the canonical commutation relation fix the form of the momentum operator?
For one dimensional quantum mechanics $$[\hat{x},\hat{p}]=i\hbar. $$
Does this fix univocally the form of the $\hat{p}$ operator? My bet is no because $\hat{p}$ actually depends if we are on ...
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Matrix elements of momentum operator in position representation
I have two related questions on the representation of the momentum operator in the position basis.
The action of the momentum operator on a wave function is to derive it:
$$\hat{p} \psi(x)=-i\hbar\...
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Proving that $i\hbar\frac{\partial}{\partial \mathbf{p}}$ is the operator of $\mathbf{x}$ in momentum space
How can I prove that $i\hbar\frac{\partial}{\partial \mathbf{p}}$ is the operator of $\mathbf{x}$ in momentum space?
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What is the most general expression for the coordinate representation of momentum operator?
I have a question about deriving the coordinate representation of momentum operator from the canonical commutation relation, $$[x,p]= i.$$
One derivation (ref W. Greiner's Quantum Mechanics: An ...
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Proof that the momentum operator is hermitian without assuming the wavefunction approaches zero at infinity?
I am currently taking my second semester of quantum mechanics. For a number of proofs in the course, we have used the assumption that the wavefunction goes to zero at infinity. We have simply used the ...
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Why operator of kinetic energy has a double derivative instead of square of single derivative?
I know that operator for $p = {h\over i} {d\over dx}$.
so $p = {h\over i} {d\psi\over dx}$ where $\psi$ is the wave function.
So, $T$ (kinetic energy) $ = {p^2 \over 2m} = {-h^2\over 2m} {d\psi \over ...
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What is the origin of the quantum operators for $p$ and $E$ in standard QM?
It is always stated the quantum operators for ${p}$ and $E$ are the ones we´re familiar with (the operator for energy, $E=i\hbar\frac{\partial}{\partial t}$ and the momentum operator, $p=-i\hbar\...
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Is commutation relation $[\hat x, \hat p]= i \hbar$ or momentum operator $\hat p = -i \hbar \nabla$ an axiom of QM?
In first year QM we postulated $\hat p = -i \hbar \nabla$ and used it to derive that $[\hat x, \hat p]= i \hbar$.
In fourth year QM they postulated $[\hat x, \hat p]= i \hbar$ and used spatial ...
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Is there any value in the following observation?
In the modern interpretation of differential geometry a tangent vector is identified as a derivation: $\frac{\partial}{\partial x}$.
In Quantum Mechanics, momentum - which is classically understood ...
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